A diver runs horizontally with a speed of 1.5 m/s off a platform that is 20 m above the water. what is his speed just before hitting the ground?

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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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I hope the diver will hit water, not ground :)

The velocity is a vector, and in this case it will be neither vertical nor horizontal. Let's find its projection to the horizontal (x) and vertical (y) axes.

Horizontally, the diver moves with the constant speed:

`x(t) = v_0*t,` where `v_0` is the initial speed (1.5m/s).

The horizontal component of the velocity is `v_0` all the time.

Vertically, the movement is uniformly accelerated:

`y(t) = h_0 - (g*t^2)/2,` where `h_0` is the initial height (20m) and g is the gravity acceleration `(9.81m/s^2).`

The vertical component of the velocity is `-g*t.` Minus sign means downward velocity.

Diver hits the water when y(t)=0 and t>0. It is `t_0 = sqrt(2*h_0/g).`

At that moment, the vertical component of the velocity will be:

`-g*t_0 = -sqrt(2*g*h_0) = -sqrt(2*9.81*20) approx -19.8 (m/s).`

So, the final velocity has horizontal component of 1.5m/s and vertical of 19.8m/s downward. The magnitude of this velocity is `sqrt(1.5^2 + 19.8^2) approx` 19.9 (m/s).